1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integration of a natural log

  1. Aug 24, 2004 #1
    [SOLVED] Integration of a natural log

    I am asked to Integrate by parts

    [tex]\int \ln(2x+1) dx[/tex]

    So,
    [tex]\mbox{u}=ln(2x+1)[/tex]
    [tex]\mbox{du}=\frac{2}{2x+1}[/tex]
    [tex]\mbox{dv}=\mbox{dx}[/tex]
    [tex]\mbox{v}=\mbox{x}[/tex]
    I plug all of that in and I get,

    [tex]{\int \ln(2x+1) dx\}={\mbox{x}\ln(2x+1)}-{\int \frac{2x}{2x+1} \mbox{dx}}[/tex]

    At this point I tried doing another substitution letting u equal 2x, but this just brought me back to the original question.
    After this anything I try to do doesn't lead me anywhere. I know it can be solved because the answer is in the back of the book.
    First, is what I have done so far correct?
    Second, how do I finish the problem?
    Any help you have would be greatly appreciated.

    Erik
     
  2. jcsd
  3. Aug 24, 2004 #2

    jamesrc

    User Avatar
    Science Advisor
    Gold Member

    Try approaching it like this:

    Let θ = 2x+1, so that your integral is:

    [tex] \int{\ln\left(2x+1\right) dx} = 2\int{\ln\theta d\theta} [/tex]

    Now you can integrate by parts like you did before, with:

    u = ln(θ) --> du = dθ/θ
    dv = dθ --> v = θ

    You should get:

    [tex] 2\left[ \theta\ln\theta - \theta \right] [/tex]

    as your result. Now all you have to do is plug θ = 2x+1 back in to find the final indefinite integral.
     
  4. Aug 24, 2004 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    jamesrc's suggestion is good. If you want to go ahead do it the way you were, you can actually do the division [itex]\frac{2x}{2x+1}= 1- \frac{1}{2x+1}[/itex]. Of course, to do the integral of [itex]\frac{1}{2x+1}[/itex], make the substition u= 2x+1 (which gets you right back to jamesrc's suggestion).
     
    Last edited: Aug 25, 2004
  5. Aug 25, 2004 #4
    thanks for your help, I hate it when the answer is so obvious and I can't figure it out.

    Thanks, Erik
     
  6. Aug 25, 2004 #5
    One final question, when you substituted theda, shouldn't the constant in front of the integral be 1/2 instead of 2???

    (using u instead of theda)

    u=2x-1
    du=2dx
    dx=(1/2)du
     
  7. Aug 25, 2004 #6
    Yes, it should be 1/2.
     
  8. Aug 25, 2004 #7

    jamesrc

    User Avatar
    Science Advisor
    Gold Member

    Yes, sorry about that. At least that proves that you know what you're doing.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Integration of a natural log
  1. Natural log (Replies: 17)

Loading...